Question 6 of 8: Torque Diagram, Maximum Shear Stress and Angle of Twist for a Stepped Shaft
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
EGBC National Exam — Basic Studies, 04-BS-6 Mechanics of Materials, 2019-Dec. Closed book; one hand-written aid sheet permitted. The exam instructs "any FIVE of the eight questions constitute a complete paper," but every question is solved below as a full study resource.
Reference texts: Hibbeler, Mechanics of Materials, 10th ed. (axial deformation of composite members ch.4; shear/moment diagrams ch.6; stress transformation & Mohr's circle ch.9; buckling of columns ch.13; torsion of circular shafts ch.5; transformed-section composite beams ch.6; combined axial+bending+shear ch.8; beam deflection by integration ch.9). Standard W-shape table (exam Appendix C) is not needed by any question below.
Question 6: Torque Diagram, Maximum Shear Stress and Angle of Twist for a Stepped Shaft (5+10+5 marks)
Applied torques: $T_B=3\text{ kN}\cdot\text{m}$, $T_C=10\text{ kN}\cdot\text{m}$, $T_D=4\text{ kN}\cdot\text{m}$. $G=80\text{ GPa}$, $\tau_Y=150\text{ MPa}$, fixed at A.
Find. (1) internal torque diagram; (2) $\tau_{max}$ and its radial distribution; (3) angle of twist of D relative to A.
Stepped shaft fixed at A; torques at B, C, D. B and C sweep the SAME rotational sense, D the opposite sense.
Check: the torque senses are taken from the arcs in the printed figure (tail vs. arrowhead position). Tracing each arc directly: $T_B$ and $T_C$ sweep bottom-of-shaft to top (arrowhead at top, SAME rotational sense); $T_D$ sweeps top to bottom (arrowhead at bottom, OPPOSITE sense to B and C). Solved below with $T_B,T_C$ acting together and $T_D$ opposing them; the resulting $\tau_{BC}=141\text{ MPa}$ sits sensibly under the 150 MPa yield (the alternative reading gives the same governing segment).
Approach. Cut the shaft in each segment and sum the applied torques on the free end to get the internal torque; then apply the torsion formula per segment and superpose the twist contributions (all rigidly connected in series, so twists ADD).
Internal torques (method of sections, free end at D). Taking the sense of $T_D$ as positive: $T_{CD}=T_D=\boxed{4\text{ kN}\cdot\text{m}}$; $T_{BC}=T_C+T_D=-10+4=\boxed{-6\text{ kN}\cdot\text{m}}$ (magnitude 6); $T_{AB}=T_B+T_C+T_D=-3-10+4=\boxed{-9\text{ kN}\cdot\text{m}}$ (magnitude 9, reacted at the wall). Diagram plotted above.
Shear stress per segment. $\tau=T(d/2)/J$: $\tau_{AB}=9\times10^6(50)/9.817\times10^6=\boxed{45.8\text{ MPa}}$; $\tau_{BC}=6\times10^6(30)/1.272\times10^6=\boxed{141.5\text{ MPa}}$; $\tau_{CD}=4\times10^6(50)/9.817\times10^6=\boxed{20.4\text{ MPa}}$. Maximum shear stress governs in segment BC (smallest diameter, moderate torque): $\tau_{max}=\boxed{141.5\text{ MPa}}<\tau_Y=150\text{ MPa}$ (elastic, as it should be for a stated yield check). The radial distribution at BC is the usual linear-in-$r$ torsion profile, zero at the axis rising to $141.5\text{ MPa}$ at the outer surface ($r=30\text{ mm}$).
Angle of twist, D relative to A. $\phi=\sum\dfrac{T_iL_i}{GJ_i}=\dfrac{(-9\times10^6)(4000)}{80{,}000(9.817\times10^6)}+\dfrac{(-6\times10^6)(4000)}{80{,}000(1.272\times10^6)}+\dfrac{(4\times10^6)(7000)}{80{,}000(9.817\times10^6)}=-0.2460\text{ rad}=\boxed{14.1^{\circ}}$ (magnitude; the sign shows D twists in the same rotational sense as B and C relative to the fixed wall at A).
